TOP 6 STEPS FOR LOG IN AND LOG OUT IN PHP

Step 1: First we have to  create database :

CREATE TABLE  `blog_login`.`user` (
`id` INT NOT NULL AUTO_INCREMENT PRIMARY KEY ,
`username` VARCHAR( 200 ) NOT NULL ,
`password` VARCHAR( 200 ) NOT NULL
) ENGINE = MYISAM

Step 2: Now create config.php : 

<?php
$host="localhost";
$user="root";
$password="";
$con=mysql_connect($host,$user,$password) or die("mysql_error()");
mysql_select_db("database_name",$con);
?>

Step 3: Now create login form :

Login.html

<form action="thanks_login.php" method="post">

<label>Username</label>
<br />
<input type="text" name="username" /><br />
<label>Password</label><br />
<input type="password" name="password" /><br />

<input type="submit" name="submit" value="Register" />
</form>

Step 4: thanks_login.php

<?php
include('config.php');
if(isset($_POST['submit'])=='Login')
{
$username=$_POST['username'];
$password=$_POST['password'];
$sql=mysql_query("select username,password from users where username='$username' and password='$password'");
if(mysql_num_rows($sql)>0)
{
session_start();
$row=mysql_fetch_array($sql);

$_SESSION['id']=$row['id'];

$_SESSION['logged-in']=true;

header("Location:welcome.php");

}
else
{
echo "invalid login";


}
?>

Step 5: now we create welcome.php

<?php
include('config.php');
session_start();
$id=$_SESSION['id'];
$sql =mysql_query("select username from user where id='$id'");
$row=mysql_fetch_array($sql);
?>

welcome <?php echo $row['username'];?><br>

<br/>
<a href="logout.php">Logout</a>

Step 6: Now create logout.php

<?php
session_start();
session_destroy();
 header("Location:index.php");
?>